Home www.play-hookey.com Tue, 08-22-2017
| Direct Current | Alternating Current | Semiconductors | Digital | Logic Families | Digital Experiments | Computers |
| Analog | Analog Experiments | Oscillators | Optics | HTML Test |
| The Fundamentals | Resistance and Reactance | Filter Concepts | Power Supplies |
| Filter Basics | Radians | Logarithms | Decibels | Low-Pass Filters | High-Pass Filters | Band-Pass Filters |

Low-Pass Filters

The Circuits

Low pass RC filter.

The two circuits to the right behave the same way for the most part, but there are some important differences. Both circuits operate as low-pass filters. That is, they will readily transmit signals below a certain frequency from input to output, with no appreciable loss in signal amplitude.

The first circuit shows a resistor and a capacitor, connected as a voltage divider. We have already looked at the behavior of this circuit with a single input frequency. The question now is, how does it behave for a range of frequencies?

One thing we can note at once: at very low frequencies, XC will be quite large, so C will have no practical effect on the signal. In essence, there will only be R in series with the signal, between vIN and vOUT. At the same time, C does provide dc isolation from ground. Sometimes this is important.

Low pass RL filter.

Our second circuit on the right is a low-pass RL filter. At low frequencies, the series inductance has negligible effect on the signal, and we essentially have a resistor connecting the signal line to ground. This is often designated a shunt resistance, indicating that it is in parallel with the signal.

As the frequency increases, XL also increases, thus increasing the total impedance of the filter. Since R doesn't change, more and more of the signal is dropped across L at higher frequencies, leaving less and less across R and available at vOUT.

At audio frequencies, the RC filter is generally preferred. This has nothing to do with any inherent benefit or lack of one circuit over the other, but rather is strictly a question of cost. Inductors for the audio frequency band are generally physically large and have heavy iron cores. This makes them expensive, and also causes them to take up a lot of space inside the cabinet of the device using it. In addition, inductors can be affected by external magnetic fields, such as the fields surrounding all power lines in homes and offices. Capacitors are much smaller, lighter, cheaper, and less susceptible to external energies, so they are used far more often.



Frequency Response

Normalized frequency response of a low pass filter.

Regardless of which circuit you choose for your filter, its frequency response will match the curve to the right. Note that the frequency scale is logarithmic rather than linear. In addition, the attenuation of the filter is shown in units called decibels (db). This is also a logarithmic unit used to show ratios. Mathematically, voltage gain or loss is described as:

Gain or Attenuation (db) = 20 log10 vOUT

vIN

Because for these filters vOUT is always less than (or equal to) vIN, the ratio is a fraction no higher than 1, so the logarithm of this fraction will be negative or zero. Thus, a voltage loss is expressed as negative decibels. A voltage gain would be expressed in positive decibels.

Regardless of the component values we use in our filter, there will be one particular frequency at which R = XC (for the RC filter), or R = XL (for the RL filter). At this frequency, vOUT/vIN = 0.707, and the calculation above will indicate a ratio of -3 db. This point is shown clearly on the graph. At lower frequencies, attenuation is reduced and rapidly becomes 0 db. At higher frequencies, attenuation increases rapidly until it falls off at a constant slope of 20 db per frequency decade as shown.

The frequency at which the two components have equal individual impedances (without regard to phase shift) is called the cut-off frequency of that filter, designated fco or ωco, according to whether we have specified frequency in Hz or radians/second. Since this graph is normalized, that frequency is shown as 1 (actually 1 times the actual cut-off frequency). The Frequency axis on the graph thus indicates frequencies in powers of 10 above and below the cut-off frequency. The shape of the graph will remain the same for any first-order RC or RL low-pass filter of the types described on this page.

Frequencies below the cut-off frequency will be passed through without significant loss, while frequencies above the cut-off frequency will be attenuated significantly. This in fact is how the filter gets its name: it passes low frequencies but not high frequencies.

If the constant slope is extended up to the 0 db line as shown in green in the graph, the intersection point is once again the cut-off frequency. In many circumstances it is useful and easier to simply use the straight lines to describe a filter's performance. This is an acceptable approximation, and the actual curve is understood to be present.



Phase Response

Voltage vectors for first-order low-pass filters.

For a first-order low-pass filter, vOUT always lags vIN by some phase angle betweeen 0 and 90°. The vector diagrams to the right show why.

The RL vector diagram at the top properly shows that vL leads vR by 90°. However, the input voltage is applied to the series combination of the two components. Thus, vL leads vIN, while vR lags vIN. Since the output is taken from across R, vOUT = vR and lags behind vIN by some phase angle that depends on frequency.

In the case of the RC low-pass filter, vC lags vR, and again vIN is applied to the series combination of the two components. This time, however, the output is taken from across C, so vOUT = vC, which again lags vIN by some angle in the range 0 to 90°, according to the signal frequency.

Phase response of a first-order low-pass filter.

Most of the variation in phase angle occurs within one decade of the cutoff frequency, as shown in the figure to the right. Note that at frequencies well below cutoff, there is essentially no phase shift. The phase lag begins to become significant about a decade below ωco, and reaches 45° (or π/4 radians) at ωco itself. Above ωco, the output phase continues to change rapidly during the first decade, by which time the phase lag is close to 90° (π/2 radians).

Because of this shifting phase lag, non-sinusoidal signals with frequency components near ωco will be distorted by this filter. Such distortion must be taken into account when designing filters of this type for some kinds of applications.



The Calculations

We perform our calculations without considering the loading effects of any circuit connected to vOUT. Any such loading effects are calculated separately. Therefore, signal current is the same through both components in the filter, and the ratio vOUT/vIN is equal to the ratio ZOUT/ZIN.

In the case of the RC filter, ZOUT = XC. For the RL filter, ZOUT = R. On the input side, Z is always the series combination of the two elements. We'll use the RC filter as our main example here; you'll find that the RL filter works out to the same result.

We are using normalized values for these calculations. This means that we arbitrarily assign values as follows:

For convenience, we will use frequency in radians/second, symbolized by the Greek letter ω (omega), rather than frequency in Hz. You need only remember when we're done with our calculations that ω = 2πf. But for the calculations themselves, this allows us to use XC = 1/ωC and avoid performing lots of calculations using π. This simplifies many calculations, and still allows the circuit to be scaled to any impedance and any frequency.

We begin at the cut-off frequency, since that is what controls our component values. We know that R = XC at this frequency, so we have:

R  =  XC
 
   =  1

ωC
 
 
ω  =  1

RC

Since R = 1Ω and C = 1f, we can immediately calculate that ω = 1 radian/second. Then, at the cut-off frequency we have:

XC  =  1      

ωC
 
   =  1  

1
 
   =  1 Ω
 
 
Z  =  (R² + XC²)½
 
   =  (1² + 1²)½
 
   =  Ω

Thus, at the cut-off frequency (generally designated ωco), XC/Z = 1/ = 0.707 as we determined above. To calculate the vOUT/vIN ratio at all frequencies, we leave ω as a variable. We can also make it complete with R and C accounted for, as follows:

vOUT  =  XC  


vIN Z
 
   =  1/ωC

(R² + 1/(ωC)²)½
 
   =  1

ωC(R² + 1/(ωC)²)½
 
   =  1

((RωC)² + 1)½

If we use the values R = 1Ω and C = 1f as we did above, and let ω vary over a range from 0.01 to 100 radians/second, and then convert the resulting ratios to db, we will get exactly the curve shown in the graph above, in blue. In fact, that's how this graph was computed. Many years of experiments have shown that it is valid, and accurately portrays the frequency response of a real low-pass filter.

The RL low-pass filter may be computed in the same way, except that here the ratio vOUT/vIN = R/Z, and Z = (R² + (ωL)²)½. The final formula is very similar, except that we must replace (RωC)² with (ωL/R)². For the RL low-pass filter, ωco = R/L. Thus, if we pick the normalized values of R = 1Ω and L = 1H, we will get exactly the same curve as we plotted above for the RC filter.

The phase response for both filters near ωco is exactly the same, and may be calculated as:

φ  =  arctan - R  =  arctan - XL


XC R
 
   =  arctan - ωRC  =  arctan - ωL

R

Prev: Decibels Next: High-Pass Filters

All pages on www.play-hookey.com copyright © 1996, 2000-2015 by Ken Bigelow
Please address queries and suggestions to: webmaster@play-hookey.com